Thread: Simple if/else statement problem

I have to create a program that will read integer numbers from the user, one at a time. It will print these numbers out after reading them. It will continue to read numbers until it receives the number 7 or 11. If the number is both even and a multiple of 5, then it will print the number multiplied by itself. If the number is instead a multiple of 6, but is not a multiple of 5, then print twice the number (instead of the number itself).

given the following inputs, the outputs should be: (these are just examples, the number should be printed out right after receiving the input number
Input Output
3 3
10 100
12 24
15 15
7

I wrote the following code:

Code:

#include <stdio.h>

int main()
{
    int x, y;
    
    printf("Enter in a number: \n");
    scanf("%d", &x);
    
    
    if (x % 2 == 0)
    
    {
    printf("%d\n", y = x * x);
    }
    
    else
    {
        printf("%d\n", y);
    }
    
    return 0;
}

But the problem I am having is if the number is instead a multiple of 6, but is not a multiple of 5, then print twice the number (instead of the number itself). If it's better to use while loops can someone please let me know how. I have not learned while loops yet.

Any help is appreciated, but please explain.

>> "If the number is both even and a multiple of 5"

ok

Code:

if(num%2==0 && num%5==0)
{
  cout num*num
}
else  if(!(num%6) && num%5)
{
    cout << num*2 //?
}

//will try it now

EDIT:

anyways to read the number..you need a loop

so while(num!=7 && num !=11)
you need to read control structures

You can nest if statements:

Code:

if(x is even)
{
  if(x is multiple of 5)
    print x * x
  else
  {
    if(x is multiple of 6)
      print x * 2
  }
}

Originally Posted by Eman

>> "If the number is both even and a multiple of 5"

ok

Not divisible by 5, a multiple of 5 (e.g. 10, 20, etc.)

Originally Posted by itsme86

Not divisible by 5, a multiple of 5 (e.g. 10, 20, etc.)

yeah..they're divisible by 5? (and even).. :S

the while loop is basically the same as an if..except it is repetitive..and always as long as the condition is true...
so

Code:

while(num!=7 && num!=11)
{
   cin >> num ;
}

if num is not equals to 7 && not equals to 11 is the same as
while(1 && 1)

if it is was 7 and 10 say..
it would be while(1 && 0) and the loop would end


if it was while(num!=7 || num!=11)
and num was 7
it would be while(0 || 1)
same as while(1)...

EDIT:

to read: http://www.learncpp.com/cpp-tutorial...ays-and-loops/

It sounds like you're going to have to use operators && (boolean AND operator) and || (boolean OR operator). This should do the trick.

Code:

#include <stdio.h>

int main() {
	int x;
	while (x != 7 && x != 11) {
		/* While x is not equal to 7 and x is not equal to 11... */
		printf("Enter a number: ");
		scanf("%d", &x);
		if ((x % 2 == 0) && (x % 5 == 0))
			/* If x is an even number and x is a multiple of 5... */
			printf("%d\n", (x * x));
		else if (x % 6 == 0)
			/* If x is a multiple of 6... */
			printf("%d\n", (x * 2));
		else
			printf("%d\n", x);
	}
	return 0;
}

oh crap!..just realized I was using C++ code...sorry!

Originally Posted by Babkockdood

It sounds like you're going to have to use operators && (boolean AND operator) and || (boolean OR operator). This should do the trick.

Code:

#include <stdio.h>

int main() {
	int x;
	while (x != 7 && x != 11) {
		/* While x is not equal to 7 and x is not equal to 11... */
		printf("Enter a number: ");
		scanf("%d", &x);
		if ((x % 2 == 0) && (x % 5 == 0))
			/* If x is an even number and x is a multiple of 5... */
			printf("%d\n", (x * x));
		else if (x % 6 == 0)
			/* If x is a multiple of 6... */
			printf("%d\n", (x * 2));
		else
			printf("%d\n", x);
	}
	return 0;
}

lol you had to answer it for him ..lol I'm sure he "learnt" it! :P

btw i think the else if is wrong
if x = 30? then it is a multiple of 6 and 5

Code:

else if (x % 6 == 0 && (x%5))
			/* If x is a multiple of 6... */
			printf("%d\n", (x * 2));

i think should be it...

Originally Posted by Babkockdood

The OP didn't mention a condition for if a number is a multiple of six and a multiple of five. Just one for if the number is even and a multiple of five, and one for if the number is a multiple of six.

post by OP:

If the number is instead a multiple of 6, but is not a multiple of 5,

I dunno..